Question

Let $f$ be a function satisfying $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y$ and $f\left(0\right)=f^{\prime }\left(0\right)=1$ then

• A. $f$ is differentiable for all x
• B. $f^{\prime }\left(x\right)=f\left(x\right)$
• C. $f\left(x\right)=e^x$
• D. $f$ is continuous for alI x

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f$ is differentiable for all x
B $f^{\prime }\left(x\right)=f\left(x\right)$
C $f\left(x\right)=e^x$
D $f$ is continuous for alI x

We have, $f(x+y)=f\left(x\right)f\left(y\right)...........\left(1\right)$

Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)f\left(h\right)-f\left(x\right)}{h}$ using (1)

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x\right)\left[f\right(h)-1)}{h}=f\left(x\right)\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=f\left(x\right)f^{\prime }\left(0\right)=f\left(x\right)$

Integrating we get, $\log \left(f\right(x\left)\right)=x+c$

Since at x$=0,f\left(x\right)=1\Rightarrow$ we have, $c=0$

Hence $f\left(x\right)=e^x$, which is continuous and differentiable for all $x$